Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Sample Input
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
Sample Output
NO YES
#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define MAX 6000
using namespace std;
struct node
{
int x,y,w;
} E[MAX];
int dis[MAX];
int N,M,W;
int top;
bool Bellman_ford()
{
memset(dis,INF,sizeof(dis));
dis[1] = 0;
bool flag;
for(int i=1; i<N; i++)
{
flag = false;
for(int j=0; j<top; j++)
{
if(dis[E[j].x] < INF && dis[E[j].y] > dis[E[j].x] + E[j].w)
{
flag = true;
dis[E[j].y] = dis[E[j].x] + E[j].w;
}
}
if(!flag)
break;
}
for(int j=0; j<top; j++)
{
if(dis[E[j].x] < INF && dis[E[j].y] > dis[E[j].x] + E[j].w)
{
return true;
}
}
return false;
}
int main()
{
// freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
top = 0;
scanf("%d%d%d",&N,&M,&W);
int a,b,c;
for(int i=0; i<M; i++)
{
scanf("%d%d%d",&a,&b,&c);
E[top].x = a;
E[top].y = b;
E[top++].w = c;
E[top].x = b;
E[top].y = a;
E[top++].w = c;
}
for(int i=0; i<W; i++)
{
scanf("%d%d%d",&a,&b,&c);
E[top].x = a;
E[top].y = b;
E[top++].w = -c;
}
if(Bellman_ford())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}